Computational-basis measurement weights are squared amplitudes

By [F1] and [F2], $P_y\mathrm{\mid }z⟩={\delta }_{y,z}\mathrm{\mid }y⟩$; hence the $P_y$ are pairwise orthogonal self-adjoint idempotents and sum to the identity, as can be checked on every basis vector. For $z={\sum }_w{b}_w\mathrm{\mid }w⟩$, $⟨P_{y}z,z{⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=\mathrm{\mid }{b}_y{\mathrm{\mid }}^2\ge 0$, so these projections are POVM effects by [F4]–[F5]. Also $⟨y,\psi {⟩}_{\mathrm{r}\mathrm{o}\mathrm{o}\mathrm{t}}=\overline{a_y}$, so $\rho P_y\mathrm{\mid }z⟩=0$ for $z\mathrm{\ne }y$ and $\rho P_y\mathrm{\mid }y⟩=\overline{a_y}\psi$. The coefficient of $\mathrm{\mid }y⟩$ in this last vector is $\mathrm{\mid }{a}_y{\mathrm{\mid }}^2$. Thus the only nonzero diagonal entry of $\rho P_y$ in this basis is $\mathrm{\mid }{a}_y{\mathrm{\mid }}^2$, and [F6] gives $\mathrm{tr}(\rho P_y)=\mathrm{\mid }{a}_y{\mathrm{\mid }}^2$.

The vector projector $\rho$ is a density operator by [F3]. The computational-basis projective measurement and its associated POVM law [F4]–[F5] therefore assign weight $\mathrm{tr}(\rho P_y)=\mathrm{\mid }{a}_y{\mathrm{\mid }}^2$ to outcome $y$. Since the basis is orthonormal and $\mathrm{\parallel }\psi \mathrm{\parallel }=1$, ${\sum }_y\mathrm{\mid }{a}_y{\mathrm{\mid }}^2=1$; each term is nonnegative. Hence the weights form a probability distribution.